Let $\Omega\subseteq\mathbb{R}^n$ be a bounded domain with smooth boundary, $n\geq2$. Schauder regularity theory says:
If $u\in H_0^1(\Omega)$ satisfies $-\Delta u=f$, with $f\in C^{\alpha}(\bar{\Omega})$, $0<\alpha<1$, then $u\in C^{2,\alpha}(\bar{\Omega})$.
I want to ask about the following attempt of proof. Extend $f$ to $\mathbb{R}^n$ by $f|_{\mathbb{R}^n\backslash\bar{\Omega}}=0$. Consider the fundamental solution $$\Phi(x)=C\log|x|,\;\;n=2;\quad\quad \Phi(x)=C\frac{1}{|x|^{n-2}},\;\;n\geq3.$$ Let $$v(x)=\int_\Omega\Phi(x-y)f(y)\,dy=(\Phi\ast f)(x),\;x\in\mathbb{R}^n.$$ The idea is to show that $v\in C^{2,\alpha}(\bar{\Omega})$ from $f\in C^{\alpha}(\bar{\Omega})$.
My questions:
Why does proving that for $v$ give $u\in C^{2,\alpha}(\bar{\Omega})$?
Trying to prove $v\in C^1(\bar{\Omega})$, if we derivate under the integral sign we have $$\partial_{x_i}v(x)=C\int_\Omega \frac{x_i-y_i}{|x-y|^n}f(y)\,dy.\quad (*)$$ But, how is it justified that it can be derivated under the integral sign? I see that $\frac{x_i-y_i}{|x-y|^n}\in L^1(\Omega,dy)$, but to derivate under the integral one would need to bound $|\frac{x_i-y_i}{|x-y|^n}|\leq g(y)\in L^1(\Omega,dy)$. I think $(*)$ is true in the distributional sense, but that is not useful to prove $v\in C^1$.
For the second derivatives, the same problem: formally, $$\partial_{x_i,x_j}u(x)=\int_\Omega\left(\frac{\delta_{ij}}{|x-y|^n}-\frac{n(x_i-y_i)(x_j-y_j)}{|x-y|^{n-2}}\right)f(y) \,dy,$$ but I do not see how this can be justified.
My doubts 2 and 3 came from reading from page 3 here.